Physics Q31 – JEE-MAIN 2025

Home Physics Rotational MotionQuestion #449

A force of

49 N

acts tangentially at the highest point of a sphere (solid) of mass

20 kg

, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

Clue 1 of 3: Rolling Motion Dynamics

For a body rolling without slipping, both translational and rotational motions are involved, and they are coupled.

🥷

Ninja Strategy•Direction of Friction and Magnitude Estimation

Recognize that when force is applied at the top of a sphere, friction acts forward, making the acceleration of the center of mass greater than if only the force $F$ acted ( $a > F / M$ ). This eliminates options smaller than $F / M = 2.45 {m/s}^{2}$ .

🧠 Full Solution Path

Step 1: Identify Forces and Equations of Motion✦ Active

Let $F$ be the applied force, $M$ be the mass, $R$ be the radius, and $a_{c}$ be the acceleration of the center of mass. The force $F$ is applied tangentially at the highest point. This force tends to make the point of contact with the ground slip backward relative to the ground. Therefore, the static friction force $f$ acts in the forward direction (same as $F$ ). The equations of motion are:

F + f = M a_{c} (Translational motion)

For rotational motion about the center of mass (taking clockwise as positive for angular acceleration $α$ ):

F \cdot R - f \cdot R = I α (Rotational motion)

For a solid sphere, the moment of inertia about its center of mass is $I = \frac{2}{5} M R^{2}$ . For rolling without slipping, $a_{c} = R α$ , which implies $α = \frac{a_{c}}{R}$ .

Step 2: Solve the System of Equations○ Expand

Substitute $I$ and $α$ into the rotational equation:

F \cdot R - f \cdot R = (\frac{2}{5} M R^{2}) (\frac{a_{c}}{R})

F - f = \frac{2}{5} M a_{c}

Now we have a system of two linear equations for $f$ and $a_{c}$ :

1) F + f = M a_{c}

2) F - f = \frac{2}{5} M a_{c}

Adding equation (1) and (2) eliminates $f$ :

(F + f) + (F - f) = M a_{c} + \frac{2}{5} M a_{c}

2 F = (1 + \frac{2}{5}) M a_{c}

2 F = \frac{7}{5} M a_{c}

a_{c} = \frac{10 F}{7 M}

💡 Teacher's Secret Hint

Alternatively, one could take torques about the point of contact to directly eliminate friction from the torque equation.

Step 3: Calculate the Acceleration○ Expand

Substitute the given values: $F = 49 N$ and $M = 20 kg$ .

a_{c} = \frac{10 \times 49}{7 \times 20}

a_{c} = \frac{10 \times (7 \times 7)}{7 \times 20}

a_{c} = \frac{10 \times 7}{20}

a_{c} = \frac{70}{20}

a_{c} = 3.5 {m/s}^{2}

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Physics Question 31 – JEE-MAIN 2025

Clue 1 of 3: Rolling Motion Dynamics

🧠 Full Solution Path